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Which of the following is formed when mercurous nitrate reacts with \(\mathrm{NaCl}\) ? (a) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Hg}\left(\mathrm{Na}_{2}\right) \mathrm{Cl}\) (c) \(\mathrm{HgCl}_{2}\) (d) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\)

Short Answer

Expert verified
The product is \(\text{Hg}_2\text{Cl}_2\) (d).

Step by step solution

01

Identify the Chemical Formula for Mercurous Nitrate

Mercurous nitrate is chemically represented as \(\text{Hg}_2(\text{NO}_3)_2\). It consists of \(\text{Hg}_2^{2+}\) ions paired with \(\text{NO}_3^-\) ions.
02

Write the Reaction Equation with NaCl

When \(\text{Hg}_2(\text{NO}_3)_2\) reacts with \(\text{NaCl}\), displacement occurs, where \(\text{Cl}^-\) ions react with \(\text{Hg}_2^{2+}\) ions to form \(\text{Hg}_2\text{Cl}_2\). The equation can be written as: \[ \text{Hg}_2(\text{NO}_3)_2 + 2\text{NaCl} \rightarrow \text{Hg}_2\text{Cl}_2 + 2\text{NaNO}_3 \]
03

Determine the Precipitate

From the balanced equation, \(\text{Hg}_2\text{Cl}_2\) is identified as the precipitate formed due to the low solubility of mercurous chloride in water.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mercurous Nitrate
Mercurous nitrate is a chemical compound with the formula \( ext{Hg}_2( ext{NO}_3)_2\). In its name, "mercurous" refers to the presence of mercury in its +1 oxidation state. This means that two mercury atoms share a pair of electrons, forming the \( ext{Hg}_2^{2+}\) ion.
This compound is usually found in a solid crystalline form and can dissolve in water, separating into mercury ions \( ext{Hg}_2^{2+}\) and nitrate ions \( ext{NO}_3^-\). However, in solutions, mercurous nitrate is sensitive and can react with other substances, such as sodium chloride (\( ext{NaCl}\)), leading to chemical changes. Understanding these properties helps in predicting its behavior in various reactions.
Displacement Reaction
A displacement reaction occurs when an element in a compound is replaced by another element from a different compound. In our exercise, when mercurous nitrate \((\text{Hg}_2( ext{NO}_3)_2)\) reacts with sodium chloride \((\text{NaCl})\), a type of displacement occurs.
The chloride ions \((\text{Cl}^-)\) from NaCl replace the nitrate \((\text{NO}_3^-)\) ions linked with mercury in mercurous nitrate. This results in the formation of mercurous chloride \((\text{Hg}_2 ext{Cl}_2)\) and sodium nitrate \((\text{NaNO}_3)\).
This type of reaction showcases how elements can "swap places," creating new products from existing compounds. Displacement reactions are common in chemistry and help in manufacturing various compounds.
Precipitation Reaction
A precipitation reaction is a type of chemical reaction where a solid, called a precipitate, forms in a liquid solution.
In our example with mercurous nitrate and sodium chloride, the reaction product, mercurous chloride \((\text{Hg}_2 ext{Cl}_2)\), does not dissolve well in water. As a result, it forms a solid precipitate that settles out of the solution.
Such reactions are crucial in chemistry for separating substances, forming new materials, and in analytical chemistry to indicate the presence of specific ions in a solution. By understanding precipitation, chemists can predict the outcomes and conditions necessary for these reactions to occur.
Solubility
Solubility is the ability of a substance to dissolve in a solvent, such as water. Every substance has a certain solubility level, determining how much of it can dissolve at a specific temperature and pressure.
In the case of mercurous chloride \((\text{Hg}_2 ext{Cl}_2)\), it is known for its low solubility in water. This means even a small amount of the compound will quickly form a precipitate rather than dissolve.
  • Factors affecting solubility include temperature, the nature of the solvent and solute, and pressure.
  • An understanding of solubility principles helps predict whether a reaction will lead to the formation of a precipitate and is essential for fields ranging from engineering to environmental science.
By understanding solubility, students and chemists learn to manipulate conditions to achieve desired chemical processes.

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Most popular questions from this chapter

A metal gives two chlorides \(\mathrm{A}\) and \(\mathrm{B}\). A gives black precipitate with \(\mathrm{NH}_{3}\) and \(\mathrm{B}\) gives white. With \(\mathrm{KI}, \mathrm{B}\) gives a yellow precipitate, which is soluble in excess of \(\mathrm{KI} . \mathrm{A}\) and \(\mathrm{B}\) are respectively (a) \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, \mathrm{HgCl}_{2}\) (b) \(\mathrm{HgCl}, \mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (c) \(\mathrm{ZnCl}_{2}, \mathrm{HgCl}_{2}\) (d) \(\mathrm{HgCl}_{2}, \mathrm{ZnCl}^{2}\)

Which of the following is formed when silver nitrate is treated with caustic soda? (a) \(\mathrm{Ag}_{3} \mathrm{~N}\) (b) \(\mathrm{Ag}\) (c) \(\mathrm{AgNO}_{2}\) (d) \(\mathrm{Ag}_{2} \mathrm{O}\)

The number of moles of \(\mathrm{AgCl}\) precipitated when excess \(\mathrm{AgNO}_{3}\) is mixed with one mole of \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{4}\right.\) \(\left.\mathrm{Cl}_{2}\right] \mathrm{Cl}\) is (a) 4 (b) 3 (c) 2 (d) 1

Cuprous ion is colourless, while cupric ion is coloured, because (a) cuprous ion has incomplete d orbital and cupric ion has a complete \(\mathrm{d}\) orbital (b) cuprous ion has a complete d orbital and cupric ion has incomplete d orbital (c) cuprous ion has half filled d orbital (d) cupric ion has unpaired electrons in d orbital

A certain metal \(\mathrm{A}\) is boiled in dilute nitric acid to give a salt B and an oxide of nitrogen \(\mathrm{C}\). An aqueous solution of \(\mathrm{B}\) with brine gives a precipitate \(\mathrm{D}\) which is soluble in \(\mathrm{NH}_{4} \mathrm{OH}\). On adding aqueous solution of \(\mathrm{B}\) to hypo solution, a white precipitate \(\mathrm{E}\) is obtained. \(\mathrm{E}\) turns black on standing. Identify A and E here? (a) \(\mathrm{Zn}, \mathrm{Zn}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{Ag}, \mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) (c) \(\mathrm{Ag}, \mathrm{Ag}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{Fe}, \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\)

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